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x=4/2x-5=3/5
We move all terms to the left:
x-(4/2x-5)=0
Domain of the equation: 2x-5)!=0We get rid of parentheses
x∈R
x-4/2x+5=0
We multiply all the terms by the denominator
x*2x+5*2x-4=0
Wy multiply elements
2x^2+10x-4=0
a = 2; b = 10; c = -4;
Δ = b2-4ac
Δ = 102-4·2·(-4)
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{33}}{2*2}=\frac{-10-2\sqrt{33}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{33}}{2*2}=\frac{-10+2\sqrt{33}}{4} $
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